如何使用一行命令同时解析多个文件(Linux/Unix)

Question

我需要解析的日志文件很多，下面是示例结构。我需要用一行命令（Linux/Unix）同时查看每个目录下的debug.log个文件内容。可能与 awk/sed/cat 等。目录数量（1、2、3、4 ...）各不相同。我需要用名称而不是“*”来限制目录。

1/logs/debug.log
2/logs/debug.log
3/logs/debug.log
4/logs/debug.log
5/logs/debug.log
6/logs/debug.log

如果每个debug.log内容如下：

In 1/logs/debug.log: finished. 
               Please go ahead next step.     
In 2/logs/debug.log: failed.      
In 3/logs/debug.log: finished. 
               Please go ahead next step.     
In 4/logs/debug.log: finished. 
               Please go ahead next step.     
In 5/logs/debug.log: Error.       
In 6/logs/debug.log: finished. 
               Please go ahead next step.

预期产出：

1: finished. 
1: Please go ahead next step.     
2: failed.    
3: finished. 
3: Please go ahead next step.     
4: finished. 
4: Please go ahead next step.     
5: Error.     
6: finished. 
6: Please go ahead next step.

对于非常复杂的问题，我们深表歉意，但如果您能告诉我解决方案的方向，我们将不胜感激。

我得到了答案

需要使用'Loop'遍历所有子目录

Answer 1

awk -v OFS=': ' 'FNR==1{dir=FILENAME; sub("/.*","",dir)} {print dir, [=10=]}' */logs/debug.log

如何使用一行命令同时解析多个文件(Linux/Unix)

How can I parse many files at the same time with one line commands(Linux/Unix)

linux

awk

cat

我得到了答案